I correct a little typo in my last answer.
With your notation:
OP:PH = k:1-k
you mean:
OP = k OH
so P = G for k=1:3 (and not for k=-2!)
Friendly
François

>You are right.
>I don't see the link between your
>question and the answer of M.Rousee
>except both are about affine geometry.

Yes, it's an error of mine.
I have forgotten to change the subject.

Etienne




[Non-text portions of this message have been removed]

Dear Jeff
> Given circlecevian triangle A'B'C' of a point P, I know at least
one
> point whose circlecevian triangle is nested within A'B'C' (incsribed
in
> A'B'C').  But, as such, there must exist at least one other point
whose
> circlecevian triangle circumscribes A'B'C.  Does anyone know this
> particular circlecevian triangle and its construction?

If T(P) is the circlecevian triangle of P, there exists one and only
one point Q such as T(Q) is inscribed in T(P) : T(Q) is the cevian
triangle of P* wrt T(P) (P* = isogonal conjugate of P wrt ABC) and, of
course, Q is the perspector of ABC and T(Q).
If I well understand, you want to inverse the transformation P->Q; it
doesn't seem so easy because, for a given point Q, we can get 4 points
P but I think that they are ruler and compass constructible although
I've no idea for the construction.
Kind regards, sincerely. Jean-Pierre

Dear Jeff and Jean-Pierre,

> [JB] Given circlecevian triangle A'B'C' of a point P, I know at least
> one
> > point whose circlecevian triangle is nested within A'B'C' (incsribed
> in
> > A'B'C').  But, as such, there must exist at least one other point
> whose
> > circlecevian triangle circumscribes A'B'C.  Does anyone know this
> > particular circlecevian triangle and its construction?
>
> [JPE] If T(P) is the circlecevian triangle of P, there exists one
> and only
> one point Q such as T(Q) is inscribed in T(P) : T(Q) is the cevian
> triangle of P* wrt T(P) (P* = isogonal conjugate of P wrt ABC) and, of
> course, Q is the perspector of ABC and T(Q).
> If I well understand, you want to inverse the transformation P->Q; it
> doesn't seem so easy because, for a given point Q, we can get 4 points
> P but I think that they are ruler and compass constructible although
> I've no idea for the construction.

I had found the same results and a strong connection with circular
isogonal pK.

All these points P, P*, Q, Q*, the vertices of T(P), T(Q) lie on the
isogonal pK with pivot the infinite point of the line PP*.
I have found several other points on the pK.

Now given Q ( and therefore Q* and T(Q) ), the sought points are the
intersections of the pK with the polar conic of Q* in this pK.
This gives a (easy to construct) conic containing the points.

This shows that the (inverse) problem does not always have solutions.


Best regards

Bernard



[Non-text portions of this message have been removed]

>From: Thomas Schoch <Thomas.Schoch@...>
>Subject: My Arbelos pages
>Date: Fri, 1 Jul 2005 21:17:34 +0200
>
>
>Dear Arbelos friends,
>
>I happily announce to you the Arbelos pages at
>my homepage (www.retas.de):
>
>  http://www.retas.de/thomas/arbelos/arbelos.html
>
>Any comments are welcome. It would be nice if
>you put a link to it on your website using the
>URI above!
>
>If your website is part of my "references" please
>check out whether I linked to it correctly. If you
>know an Arbelos related website that I should add
>to my references, please let me know.
>
>Hope you are fine!
>
>Best regards.
>
>         Thomas Schoch
>
--

Dear Jean-Pierre and Bernard,

Thank you very much for your responses.  I did not expect the inverse
problem to be quite so involved.  Bernard, your analysis and
construction will take me some time to digest.  I would like to see
one possible solution as an example.

Following the notation below and taking P=X6 then Q has search value
equal to -0.50949614 which is not listed in ETC.  If I understand
correctly, four points can be found for the inverse problem but only
one of these will give a circlecevian triangle that will circumscribe
T(P).  Could someone provide the coordinate values of this point?
Hopefully, the choice of P=X6 gives a solution.

Thank you,
Jeff Brooks


> > [JB] Given circlecevian triangle A'B'C' of a point P, I know at
least
> > one
> > > point whose circlecevian triangle is nested within A'B'C'
(incsribed
> > in
> > > A'B'C').  But, as such, there must exist at least one other
point
> > whose
> > > circlecevian triangle circumscribes A'B'C.  Does anyone know
this
> > > particular circlecevian triangle and its construction?
> >
> > [JPE] If T(P) is the circlecevian triangle of P, there exists
one
> > and only
> > one point Q such as T(Q) is inscribed in T(P) : T(Q) is the cevian
> > triangle of P* wrt T(P) (P* = isogonal conjugate of P wrt ABC)
and, of
> > course, Q is the perspector of ABC and T(Q).
> > If I well understand, you want to inverse the transformation P-
>Q; it
> > doesn't seem so easy because, for a given point Q, we can get 4
points
> > P but I think that they are ruler and compass constructible
although
> > I've no idea for the construction.
>
> I had found the same results and a strong connection with circular
> isogonal pK.
>
> All these points P, P*, Q, Q*, the vertices of T(P), T(Q) lie on
the
> isogonal pK with pivot the infinite point of the line PP*.
> I have found several other points on the pK.
>
> Now given Q ( and therefore Q* and T(Q) ), the sought points are
the
> intersections of the pK with the polar conic of Q* in this pK.
> This gives a (easy to construct) conic containing the points.
>
> This shows that the (inverse) problem does not always have
solutions.
>
>
> Best regards
>
> Bernard
>
>
>
> [Non-text portions of this message have been removed]

Dear Jeff and Bernard,
may be this is more clear.
For a given point, let
M* = isogonal conjugate of M
T(M) = circlecevian triangle of M with vertices a(M),b(M),c(M)
Now, for a given point P, there exists an unique point Q such as T
(Q) is inscribed in T(P) : T(Q) is the cevian triangle of P* wrt T(P)
There exists, apart P, three other points M such as T(Q) is
inscribed in T(M) : the vertices of T(P*) (they are the isogonal
conjugates of the vertices of T(P))
For instance, T(a(P*)) = P*c(P)b(P),...
If we take P = K, Q is barycentric
x = (b^2+c^2+4a^2)/(a^4+b^4+c^4+11b^2c^2-7a^2(b^2+c^2))
and T(Q) is inscribed in T(K), T(a(G)), T(b(G)), T(c(G)) where a(G)b
(G)c(G) = circlecevian triange of G.
For instance a(G) : -3a^2/(a^2+b^2+c^2):1:1

It follows that, for a given point M, there exists generally 0 or 4
points P such as T(P) circumscribes T(M). All these points are ruler
and compass constructible (but how to do?).
In the case M = K, there exists no point P such as T(P)
circumscribes T(K).
Of course, it could be interesting to look at the set of the points
M for which there exist 4 points P such as T(P) circumscribes T(M).
Kind regards. Jean-Pierre

> Thank you very much for your responses.  I did not expect the
inverse
> problem to be quite so involved.  Bernard, your analysis and
> construction will take me some time to digest.  I would like to
see
> one possible solution as an example.
>
> Following the notation below and taking P=X6 then Q has search
value
> equal to -0.50949614 which is not listed in ETC.  If I understand
> correctly, four points can be found for the inverse problem but
only
> one of these will give a circlecevian triangle that will
circumscribe
> T(P).  Could someone provide the coordinate values of this point?
> Hopefully, the choice of P=X6 gives a solution.
>
> Thank you,
> Jeff Brooks
>
>
> > > [JB] Given circlecevian triangle A'B'C' of a point P, I know
at
> least
> > > one
> > > > point whose circlecevian triangle is nested within A'B'C'
> (incsribed
> > > in
> > > > A'B'C').  But, as such, there must exist at least one other
> point
> > > whose
> > > > circlecevian triangle circumscribes A'B'C.  Does anyone know
> this
> > > > particular circlecevian triangle and its construction?
> > >
> > > [JPE] If T(P) is the circlecevian triangle of P, there exists
> one
> > > and only
> > > one point Q such as T(Q) is inscribed in T(P) : T(Q) is the
cevian
> > > triangle of P* wrt T(P) (P* = isogonal conjugate of P wrt ABC)
> and, of
> > > course, Q is the perspector of ABC and T(Q).
> > > If I well understand, you want to inverse the transformation P-
> >Q; it
> > > doesn't seem so easy because, for a given point Q, we can get
4
> points
> > > P but I think that they are ruler and compass constructible
> although
> > > I've no idea for the construction.
> >
> > I had found the same results and a strong connection with
circular
> > isogonal pK.
> >
> > All these points P, P*, Q, Q*, the vertices of T(P), T(Q) lie on
> the
> > isogonal pK with pivot the infinite point of the line PP*.
> > I have found several other points on the pK.
> >
> > Now given Q ( and therefore Q* and T(Q) ), the sought points are
> the
> > intersections of the pK with the polar conic of Q* in this pK.
> > This gives a (easy to construct) conic containing the points.
> >
> > This shows that the (inverse) problem does not always have
> solutions.
> >
> >
> > Best regards
> >
> > Bernard
> >
> >
> >
> > [Non-text portions of this message have been removed]

Dear Jean-Pierre and Jeff,

> [JPE] It follows that, for a given point M, there exists generally
> 0 or 4
> points P such as T(P) circumscribes T(M). All these points are ruler
> and compass constructible (but how to do?).

How to be sure they are ruler and compass constructible ?

In this case, it would be possible to construct the common points of
a circular isogonal pK with the polar conic of one of its points !


Best regards

Bernard



[Non-text portions of this message have been removed]

Dear Jean-Pierre,

> Of course, it could be interesting to look at the set of the points
> M for which there exist 4 points P such as T(P) circumscribes T(M).

these are the points such that M lies on the infinite branch of the
isogonal circular pK with pivot the infinite point of the line MM*.

for example, with M = X1263, the pK is Neuberg and the 4 points are H
and the vertices of the reflection triangle.



Best regards

Bernard



[Non-text portions of this message have been removed]

Dear Bernard
[JPE]
> > Of course, it could be interesting to look at the set of the points
> > M for which there exist 4 points P such as T(P) circumscribes T(M).
[BG]
> these are the points such that M lies on the infinite branch of the
> isogonal circular pK with pivot the infinite point of the line MM*.

Yes but, except the limit cases, we can split the plane in two parts :
if M lies in the first part, there are 4 points P; if M lies in the
second part, there are 0 point P.
What are theses two parts?
Friendly. Jean-Pierre

Dear Bernard and Jeff
> > [JPE] It follows that, for a given point M, there exists
generally
> > 0 or 4
> > points P such as T(P) circumscribes T(M). All these points are
ruler
> > and compass constructible (but how to do?).
[BG]
> How to be sure they are ruler and compass constructible ?
>
> In this case, it would be possible to construct the common points
of
> a circular isogonal pK with the polar conic of one of its points !

If I well follow you, the points P such as T(P) circumscribes T(M)
are the isogonal conjugates of the common points - apart M* - of the
self isogonal cubic with pivot the infinite point of MM* and the
polar conic of M*.
If we want to find these 4 common points P1, P2, P3, P4, some simple
remarks lead to the fact that they are ruler and compass
constructible (more over, the construction should be very easy, but
I didn't find it ):
The diagonal triangle of P1P2P3P4 is the circle cevian triangle Ma,
Mb, Mc of M.
If P1P2 ^ P3P4 = Ma; P1P3 ^ P2P4 = Mb; P1P4 ^ P2P3 = Mc, we have
P1P2* ^ P2P1* = P3P4* ^ P4P3* = A
P1P3* ^ P3P1* = P2P4* ^ P4P2* = B
P1P4* ^ P4P1* = P2P3* ^ P3P2* = C
and, of course, P1P1*, P2P2*, P3P3*, P4P4*, MM* parallel
Friendly. Jean-Pierre

Dear Jean-Pierre and Jeff,

> If I well follow you, the points P such as T(P) circumscribes T(M)
> are the isogonal conjugates of the common points - apart M* - of the
> self isogonal cubic with pivot the infinite point of MM* and the
> polar conic of M*.
> If we want to find these 4 common points P1, P2, P3, P4, some simple
> remarks lead to the fact that they are ruler and compass
> constructible

here's a construction of the four points :

let Q be a point and A1B1C1 its circle cevian triangle.
we seek P whose circle cevian triangle A'B'C' circumscribes A1B1C1.

let A2 = BC1 /\ CB1.
let Lb be the line through B and QA /\ CB1
let Lc be the line through C and QA /\ BC1

Lb and Lc intersect at a point which is the perpector of an in-conic
in the triangle BCA2.
draw the two tangents through A1 to this conic.
they contain the four requested points.

start over with B2 and you obtain another pair of lines though B1
which now give the 4 points.

call P* one of the points and A', B', C' the 3 others such that A1,
A', P* (etc) are collinear.

then the circle cevian triangle of P is A'B'C'.

All the mentioned points lie on the isogonal circular pK with pivot
the infinite point of QQ* and the polar conic of Q* contains A', B',
C', P*.

This gives a construction of the polar conic of any point on this
type of pK with their common points.
I will add this in Special Isocubics.

A very interesting and challenging problem indeed !

Best regards

Bernard



[Non-text portions of this message have been removed]

Dear Bernard
> > If I well follow you, the points P such as T(P) circumscribes T
(M)
> > are the isogonal conjugates of the common points - apart M* - of
the
> > self isogonal cubic with pivot the infinite point of MM* and the
> > polar conic of M*.
> > If we want to find these 4 common points P1, P2, P3, P4, some
simple
> > remarks lead to the fact that they are ruler and compass
> > constructible
>
> here's a construction of the four points :
>
> let Q be a point and A1B1C1 its circle cevian triangle.
> we seek P whose circle cevian triangle A'B'C' circumscribes A1B1C1.
>
> let A2 = BC1 /\ CB1.
> let Lb be the line through B and QA /\ CB1
> let Lc be the line through C and QA /\ BC1
>
> Lb and Lc intersect at a point which is the perpector of an in-
conic
> in the triangle BCA2.
> draw the two tangents through A1 to this conic.
> they contain the four requested points.
>
> start over with B2 and you obtain another pair of lines though B1
> which now give the 4 points.
>
> call P* one of the points and A', B', C' the 3 others such that
A1,
> A', P* (etc) are collinear.
>
> then the circle cevian triangle of P is A'B'C'.
>
> All the mentioned points lie on the isogonal circular pK with
pivot
> the infinite point of QQ* and the polar conic of Q* contains A',
B',
> C', P*.
>
> This gives a construction of the polar conic of any point on this
> type of pK with their common points.
> I will add this in Special Isocubics.
>
> A very interesting and challenging problem indeed !

Very good.
In fact, it gives (after projectivity) a construction of the common
points of a pK and the polar conic of any point lying on the pK.
We need, apart Q and Q*, another pair R, R* lying on the cubic; the
line at infinity will become the line RR* and a circle will become a
conic through R and R*.
Friendly. Jean-Pierre

--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <Jean-Pierre.Ehrmann.70@n...>
wrote:
> Dear Bernard
> > > If I well follow you, the points P such as T(P) circumscribes T
> (M)
> > > are the isogonal conjugates of the common points - apart M* - of
> the
> > > self isogonal cubic with pivot the infinite point of MM* and the
> > > polar conic of M*.
> > > If we want to find these 4 common points P1, P2, P3, P4, some
> simple
> > > remarks lead to the fact that they are ruler and compass
> > > constructible
> >
> > here's a construction of the four points :
> >
> > let Q be a point and A1B1C1 its circle cevian triangle.
> > we seek P whose circle cevian triangle A'B'C' circumscribes A1B1C1.
> >
> > let A2 = BC1 /\ CB1.
> > let Lb be the line through B and QA /\ CB1
> > let Lc be the line through C and QA /\ BC1
> >
> > Lb and Lc intersect at a point which is the perpector of an in-
> conic
> > in the triangle BCA2.
> > draw the two tangents through A1 to this conic.
> > they contain the four requested points.
> >
> > start over with B2 and you obtain another pair of lines though B1
> > which now give the 4 points.
> >
> > call P* one of the points and A', B', C' the 3 others such that
> A1,
> > A', P* (etc) are collinear.
> >
> > then the circle cevian triangle of P is A'B'C'.
> >
> > All the mentioned points lie on the isogonal circular pK with
> pivot
> > the infinite point of QQ* and the polar conic of Q* contains A',
> B',
> > C', P*.
> >
> > This gives a construction of the polar conic of any point on this
> > type of pK with their common points.
> > I will add this in Special Isocubics.
> >
> > A very interesting and challenging problem indeed !
>
> Very good.
> In fact, it gives (after projectivity) a construction of the common
> points of a pK and the polar conic of any point lying on the pK.
> We need, apart Q and Q*, another pair R, R* lying on the cubic; the
> line at infinity will become the line RR* and a circle will become a
> conic through R and R*.
> Friendly. Jean-Pierre

Hello!
Hence,
we can construct the four tangents from a point of a pK.
Bye.
Fred

Dear Jean-Pierre,

> In fact, it gives (after projectivity) a construction of the common
> points of a pK and the polar conic of any point lying on the pK.
> We need, apart Q and Q*, another pair R, R* lying on the cubic; the
> line at infinity will become the line RR* and a circle will become a
> conic through R and R*.

in fact, it is not necessary to have another pair of points on the pK
if it is defined by its pivot P and its isoconjugate P*.

to construct the polar conic of U on the pK and its 4 intersections
(apart U) with the pK (I will call them pretangentials of U), all you
need is the triangle A1B1C1 formed by the third points of the pK on
the cevians of U*, very easy to draw.

my construction is still valid.
I will write down all this in Special Isocubics when time permits.

Best regards

Bernard



[Non-text portions of this message have been removed]

Dear friends,
it is known http://mathworld.wolfram.com/AdamsCircle.html
that the parallel lines from the Gergonne point
Ge = ( 1/(s-a) , 1/(s-b), 1/(s-c) ) in barycentrics
of triangle ABC to the sides of its cevian triangle meet the
sides of ABC at six points that are lying on Adams'  circle
a circle with center the incenter of ABC.

Is it known that the same property holds also for the other
Gergonne points Ga, Gb, Gc of triangle ABC?

i.e. if Ga = (-1/s , 1/(s-b), 1/(s-c) ) in barycentrics
(Ga is the common point of the lines AA', BB', CC'
where A'B'C' is the pedal triangle of the A-excenter)
then the parallels from Ga to the sides of its cevian
triangle meet the sides of ABC at six points that are
lying on a circle with center the A-excenter of ABC.

Best regards
Nikolaos Dergiades

Dear Nikos

[ND]
>it is known http://mathworld.wolfram.com/AdamsCircle.html
>that the parallel lines from the Gergonne point
>Ge = ( 1/(s-a) , 1/(s-b), 1/(s-c) ) in barycentrics
>of triangle ABC to the sides of its cevian triangle meet the
>sides of ABC at six points that are lying on Adams'  circle
>a circle with center the incenter of ABC.
>
>Is it known that the same property holds also for the other
>Gergonne points Ga, Gb, Gc of triangle ABC?

This should be known by Conway's extraversion

BTW, and Conway's Circle is also centered at I and has
three brother-circles by extraversion.


APH
--

Dear friends,

Recent messages (many thanks to Jean-Pierre and Bernard!) on nested
circle cevian triangles led me to take a look at the class of
circular pivotal cubics (CL035 in Bernard's Cubics in the Triangle
Plane).

Empirical evidence suggests a relationship between this class of
cubics and the perspective triangles listed below.  It seems after
looking at numerous examples (CONJECTURE) that if Q is the pivot of a
particular circular cubic, then the triangle A'B'C' is perspective
with ABC if and only if P lies on the cubic.

The algorithm listed below will not function properly with points on
the line at infinity or cases when p=q.  I hope I have not made a
mistake or overlooked an obvious geometrical fact but I was quite
pleased to see that the three cases below might be related in this
way.

Sincerely,
Jeff Brooks

>
> Consider the following algorithm CPLEX(P,Q):
>
> p=P;
> q=Q;
>
> g[u_,v_,w_]:= 1/(p-u)
> h[u_,v_,w_]:= 1/(q-u)
> T[u_,v_,w_]:= (g[v,w,u] - g[w,u,v]) h[u,v,w]
> f[u_,v_,w_]:= ((w-u) T[u,v,w] v + (w-v) T[v,w,u] u)/
> 	        ((w-u) T[u,v,w] + (w-v) T[v,w,u])
>

Let A'=f[B,C,A], B'=f[C,A,B], C'=f[A,B,C] which one can easily change
to be more pair-wise accurate from a cyclic standpoint.  A'B'C' is
perspective with ABC in certain cases. It now appears they are
perspective if and only if they conform to the conjecture above.

> Other observations:  I am most interested in 3) and 4) below.
>
> 1) Circumcevian Triangle of point X:
>
> p=INV[X]  -- inverse of X in the circumcircle
> q=ISOG[X] -- isogonal conjugate of X (written as X' below)
>
> 3) Umecevian or Saragossa Triangle:
>
> p=ISOG[COM[X]] – isogonal of the complement of X
> q=X
>
> A'B'C' is always perspective with ABC at a point on the
circumcircle.
>
> 4)  I need a name for this triangle:
>
> p=X
> q=ACOM[ISOG[X]]
>
> A'B'C' is always perspective with ABC at a point on the
> circumcircle.  The perspector corresponds with the circumcircle
> points in message 6416 -- but I don't know why.
>
> X(2)==>X(111)
> X(6)==>X(111)
> X(3)==>X(74)
> X(1)==>X(106)
> etc...
>
> What is the relationship here?  Is there a nice geometric proof to
> explain this?
>
> Sincerely,
> Jeff

The following paper has been published in Forum Geometricorum. It can be
viewd at

http://forumgeom.fau.edu/FG2005volume5/FG200513index.html

The editors
Forum Geometricorum
=======================================================
Peter J. C. Moses, Circles and Triangle Centers Associated with the Lucas
Circles,
Forum Geometricorum, 5 (2005) 97--106.

Abstract: The Lucas circles of a triangle are the three circles mutually
tangent to each other externally, and each tangent internally to the
circumcircle of the triangle at a vertex. In this paper we present some
further interesting circles and triangle centers associated with the Lucas
circles.

Thanks Antreas.

[ND]
> >it is known http://mathworld.wolfram.com/AdamsCircle.html
> >that the parallel lines from the Gergonne point
> >Ge = ( 1/(s-a) , 1/(s-b), 1/(s-c) ) in barycentrics
> >of triangle ABC to the sides of its cevian triangle meet the
> >sides of ABC at six points that are lying on Adams'  circle
> >a circle with center the incenter of ABC.
> >
> >Is it known that the same property holds also for the other
> >Gergonne points Ga, Gb, Gc of triangle ABC?

[APH]
> This should be known by Conway's extraversion
>
> BTW, and Conway's Circle is also centered at I and has
> three brother-circles by extraversion.


If the point P has barycentrics (x : y : z)
and A'B'C' is its cevian triangle then the parallel line
from P to B'C' meets the sides AC, AB of triangle ABC
at the points Ab, Ac. Similarly define the points Bc,Ba, Ca, Cb.
The six points are lying on a conic with center the point
( x(y+z) : y(z+x) : z(x+y) ) in barycentrics that is the
complement of the isotomic conjugate of P.
If the point P is inside the Steiner circumellipse the conic
is an ellipse or circle for the points Go, Ga, Gb, Gc.
If the point P is on the Steiner circumellipse the conic
is a parabola.
If the point P is outside of the Steiner circumellipse the conic
is an hyperbola and if P is also on the conic
xx + yy + zz + 3xy + 3yz + 3zx = 0
the Steiner circumellipse of the anticomplement triangle of ABC
the points Ab, Bc, Ca are collinear and the points Ac, Ba, Cb
are also collinear.

Best regards
Nikolaos Dergiades

Dear friends
Beeing given 3 lines L_{1}, L_{2}, L_{3} in R^3 with equations:
x = a_{k} z + b_{k}
y = c_{k} z + d_{k}
for k= 1, 2, 3
   1° Is there a simple way to write down an equation of the ruled
quadric having these 3 lines as generators?
   2° Is it possible to write easily equations for the two systems
of generators of this quadric?
Friendly
François

If three concurrent chords of a conic are equal, then is
the conic a circle (and the chords diameters) ?

aph
--

On 6-07-05, "Antreas P. Hatzipolakis" <xpolakis@...> wrote:

>If three concurrent chords of a conic are equal, then is
>the conic a circle (and the chords diameters) ?
>
>aph

Well... not.

Now, if three concurrent chords of a circle are equal,
then the three chords are necessarily diameters.

Simple proof?

aph

Dear Friends,



The question is about Math. Magazine Problem 1171 (Feb/2005).

Suppose the cevians AA', BB', CC' of triangle ABC meet at M. If M is
inside ABC and A'B'C' is similar to ABC, then it is not difficult to see
that M must be the centroid.

But from some rough sketches, it seems that there could be M outside ABC
and making A'B'C' similar to ABC. Is this true? If yes, are these
special points in Kimberling's list?

Thanks!



Li Zhou



[Non-text portions of this message have been removed]

The chords AA', BB', CC' of the circle (O) are concurrent at M.
Let l = AA' = BB' = CC', and p the power of M wrt (O).

We have :
MA + xMA' = MB + xMB' = MC + xMC' = l
MA.MA' = MB.MB' = MC.MC' = -xp
with x = 1 if M is internal to (O), x = -1 if M is external.

The system (u + xv = l, u.v = -xp) as only one non-ordered solution.
So MA = MB = MC and MA' = MB' = MC' (or permutation) and M is the center
of (O), circumcircle of ABC and A'B'C'.

Best regards

Gilles Boutte

>Now, if three concurrent chords of a circle are equal,
>then the three chords are necessarily diameters.
>
>Simple proof?
>
>

On 7/6/05, Boutte Gilles <g_bouttefr@...> wrote:
> The chords AA', BB', CC' of the circle (O) are concurrent at M.
> Let l = AA' = BB' = CC', and p the power of M wrt (O).
>
> We have :
> MA + xMA' = MB + xMB' = MC + xMC' = l
> MA.MA' = MB.MB' = MC.MC' = -xp
> with x = 1 if M is internal to (O), x = -1 if M is external.
>
> The system (u + xv = l, u.v = -xp) as only one non-ordered solution.
> So MA = MB = MC and MA' = MB' = MC' (or permutation) and M is the center
> of (O), circumcircle of ABC and A'B'C'.
>
> Best regards
>
> Gilles Boutte
>
> >Now, if three concurrent chords of a circle are equal,
> >then the three chords are necessarily diameters.
> >
> >Simple proof?
> >
Gilles's proof is very nice. Here another one, not using directly the
notion of power:
> Let R the radius of the circle, O his center and 2l the common lenght of the
three cords. These cords are tangent to a circle of center O and radius r =
Sqrt(R^2 - l^2).
If l<>R, then r<>0, and we have a nonsense with 3 distinct tangents
from M to a circle of non zero radius.
Then l=R and we are done.
Friendly
François

Dear Antreas,
we had very nice proofs for the case of circle
but before this you wrote

> If three concurrent chords of a conic are equal, then is
> the conic a circle (and the chords diameters) ?

and maybe somebody is trying to solve this problem.
The answer is no because we can construct an ellipse
that has three equal concurrent chords.

Does anybody can give the equation of a curve (not circle)
that has infinite equal concurrent chords?

Best regards
Nikolaos Dergiades

Here what the meaning of a curve?
If we begin with a "curve" C and we construct the conchoid C' of C wrt
O for a length l, the reunion of C and C' is a new "curve" C" with
infinitely many chords through O of length l?
We must restrict to some kind of curves, for example irreducible
algebraic curves?
Friendly
François

On 7/7/05, Nikolaos Dergiades <ndergiades@...> wrote:
> Dear Antreas,
> we had very nice proofs for the case of circle
> but before this you wrote
>
> > If three concurrent chords of a conic are equal, then is
> > the conic a circle (and the chords diameters) ?
>
> and maybe somebody is trying to solve this problem.
> The answer is no because we can construct an ellipse
> that has three equal concurrent chords.
>
> Does anybody can give the equation of a curve (not circle)
> that has infinite equal concurrent chords?
>
> Best regards
> Nikolaos Dergiades
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>

Dear François,
I had in mind the cardioide.
For example in polar coordinates (p, w)
the cardioide with equation p = 1 + cosw
has every chord AA' that passes through O
equal to 2 since
OA = 1 + cosw
OA' = 1 + cos(pi + w) = 1-cosw
AA' = OA + OA' = 2

Best regards
Nikolaos Dergiades

> -----Original Message-----
> From: Hyacinthos@yahoogroups.com
> [mailto:Hyacinthos@yahoogroups.com]On Behalf Of Francois Rideau
> Sent: Thursday, July 07, 2005 10:39 AM
> To: Hyacinthos@yahoogroups.com
> Subject: Re: [EMHL] chords
>
>
> Here what the meaning of a curve?
> If we begin with a "curve" C and we construct the conchoid C' of C wrt
> O for a length l, the reunion of C and C' is a new "curve" C" with
> infinitely many chords through O of length l?
> We must restrict to some kind of curves, for example irreducible
> algebraic curves?
> Friendly
> François
>
> On 7/7/05, Nikolaos Dergiades <ndergiades@...> wrote:
> > Dear Antreas,
> > we had very nice proofs for the case of circle
> > but before this you wrote
> >
> > > If three concurrent chords of a conic are equal, then is
> > > the conic a circle (and the chords diameters) ?
> >
> > and maybe somebody is trying to solve this problem.
> > The answer is no because we can construct an ellipse
> > that has three equal concurrent chords.
> >
> > Does anybody can give the equation of a curve (not circle)
> > that has infinite equal concurrent chords?
> >
> > Best regards
> > Nikolaos Dergiades
> >
> >
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> >
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>

Dear colleagues!
There are some interesting properties of two triangle centers.
Let P1, P2 are the Brocard points of triangle ABC, C0 is the common point of
AP1 and BP2, such that AC0=BC0, C1 is the reflection of C0 in AB, C2 is the
common point of perpendiculars to AC0 from A and BC0 from B, A1, A2, B1, B2
are defined similarly. Then the lines AA1, BB1, CC1 concur in X(262) and
AA2, BB2, CC2 in X(83).

Sincerely                                               Alexey




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